Relative humidity of the air entering a compressor

AI Thread Summary
The discussion focuses on calculating the Free Air Delivery (FAD) and relative humidity of air entering a compressor. The FAD is estimated at 7.9 m³/min under normal temperature and pressure conditions. To determine relative humidity, the user calculates the absolute humidity at the inlet and outlet conditions, resulting in a relative humidity of 38.6%. There is some confusion regarding the units of measurement, specifically the difference between cubic meters and cubic feet per minute. The conversation highlights the importance of accurate calculations and unit conversions in thermodynamic processes.
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Homework Statement


Air is drawn into a compressor at normal temperature and pressure (N.T.P.) and compressed to a pressure of 6 bar gauge. After compression the air is delivered at 1.2m3 min–1 and cooled to a temperature of 30°C, at which point condensate is collected at the rate of 2 litres per hour. Estimate the FAD (N.T.P.) of the compressor, and the relative humidity of the air entering the compressor.

Homework Equations


P1 * V1 / T1 = P2 * V2 / T2

The Attempt at a Solution



Inlet conditions
P1 = 1.013 Bar abs
T1 = 15°C = 288.15°K

Outlet conditions
P2 = 6 Bar + 1.013 bar = 7.013 bar abs
V2 = 1.2m3 min-1
T2 = 30°C = 303.15°K

P1*V1/T1 = P2*V2/T2

V1 = P2*T1V2 / P1*T2
V1 = 6 * 288.15 * 1.2 / 1.013 * 303.15

V1 = 7.9m3 min-1 FAD (NTP)


Can anyone help me with calculating the relative humidity of the air entering the compressor?
 
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Ive had a go at Calculating the Relative Humidity (RH). Have I used the correct method to calculate RH?Inlet conditions
7.9m3 min-1 * 60 = 474m3 / hour

Using a Dew-point Chart (at 1.013 bar) to find the amount of moisture present in saturated air at 15°C = 12.5 g

Saturation Quantity = 12.5g * 474m3 / hour = 5925 g / hourOutput conditions
1.2m3 min-1 * 60 = 72m3 / hour

Using a pressure Dew-point Chart (at 7.013 bar abs) to find the amount of moisture present in saturated air at 30°C = 4 g

Saturation Quantity = 4g * 72m3 / hour = 288 g / hour

Absolute Humidity = 288g + 2000g = 2288 g / hour

Relative Humidity = (2288 / 5925) * 100 = 38.6%
 
I read this as saying that hot air exits the compressor at 1.2 cu ft/min. CORRECTION: 1.2 m3/min. Then after this it is cooled to 30°C. Do you think this is what it's saying?

For an adiabatic process, I think you also have ##pv^\gamma##=const
but whether you need to apply that here I can't say. (I have not dealt with thermodynamics of fluids since I was a student.)

Have you looked at the similar threads listed at the foot of this page? There's no guarantee they are automatically correct, of couse.
 
Last edited:
NascentOxygen said:
I read this as saying that hot air exits the compressor at 1.2 cu ft/min. Then after this it is cooled to 30°C. Do you think this is what it's saying?
cubic meters per minute. Big difference.
 
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